If $(x+2)(x-3)=14$, find the product of the possible values of $x$.
Expanding the left side of the given equation, we have $x^2-x-6=14 \Rightarrow x^2-x-20=0$. Since in a quadratic  with equation of the form $ax^2+bx+c=0$ the product of the roots is $c/a$, the product of the roots of the given equation is $-20/1 = \boxed{-20}$.